Derivative of Sine of Function

Theorem

Let $u$ be a differentiable real function of $x$.

Then:

$\map {\dfrac \d {\d x} } {\sin u} = \cos u \dfrac {\d u} {\d x}$

where $\sin$ is the sine function and $\cos$ is the cosine function.

Proof

\(\ds \map {\frac \d {\d x} } {\sin u}\)

\(=\)

\(\ds \map {\frac \d {\d u} } {\sin u} \frac {\d u} {\d x}\)

Chain Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \cos u \frac {\d u} {\d x}\)

Derivative of Sine Function

$\blacksquare$

Also see

• Derivative of Cosine of Function

• Derivative of Tangent of Function
• Derivative of Cotangent of Function

• Derivative of Secant of Function
• Derivative of Cosecant of Function

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 13$: Derivatives of Trigonometric and Inverse Trigonometric Functions: $13.14$
• 1974: Murray R. Spiegel: Theory and Problems of Advanced Calculus (SI ed.) ... (previous) ... (next): Chapter $4$. Derivatives: Derivatives of Special Functions: $3$